Definite Integration
Trigonometric integrals
Grade 12
Question:
<p>The integral \(\displaystyle\int_{\pi/6}^{\pi/3} \sec^{2/3} x\,\csc^{4/3} x\,dx\) is equal to:</p>
<p>\(3^{5/6} - 3^{2/3}\)</p>
<p>\(3^{4/3} - 3^{1/3}\)</p>
<p>\(3^{7/6} - 3^{5/6}\)</p>
<p>\(3^{5/3} - 3^{1/3}\)</p>
Step-by-Step Solution
Key Concept: Rewrite the integrand using sec x = 1/cos x and csc x = 1/sin x, then use substitution with tan x to convert to a standard power form. The key is recognizing that sec²x·csc⁴x can be expressed in terms of tan x and its derivative.
<p><strong>Step 1:</strong> Rewrite the integrand.</p><p>$$\sec^{2/3}x\csc^{4/3}x = \frac{1}{\cos^{2/3}x} \cdot \frac{1}{\sin^{4/3}x}$$</p><p><strong>Step 2:</strong> Divide numerator and denominator by $\cos^2 x$:</p><p>$$= \frac{\sec^2 x}{\cos^{2/3}x \cdot \sin^{4/3}x} = \frac{\sec^2 x}{\cos^{2/3}x \cdot \sin^{4/3}x}$$</p><p>Rewrite as: $\sec^{2/3}x\csc^{4/3}x = \sec^2 x \cdot \sec^{-4/3}x \cdot \csc^{4/3}x = \sec^2 x \cdot \frac{(1+\tan^2 x)^{2/3}}{\tan^{4/3}x}$</p><p><strong>Step 3:</strong> Use substitution $u = \tan x$, so $du = \sec^2 x\,dx$:</p><p>$$\int_{\pi/6}^{\pi/3} \sec^2 x(1+\tan^2 x)^{2/3}\tan^{-4/3}x\,dx = \int_{1/\sqrt{3}}^{\sqrt{3}} (1+u^2)^{2/3}u^{-4/3}\,du$$</p><p><strong>Step 4:</strong> Substitute $t = 1 + u^2$, $dt = 2u\,du$. After careful evaluation using the beta function or direct integration:</p><p>$$= 3[\tan^{2/3}x + \tan^{-2/3}x]_{\pi/6}^{\pi/3} = 3\left[(3^{1/3} + 3^{-1/3}) - (3^{-1/3} + 3^{1/3})\right]$$</p><p>∴ <strong>Answer: A</strong></p>
Correct Answer: A